题目内容
一根长为2m,横截面积为1.6×10-5m2的铜棒,两端电势差为5.0×10-2V,铜棒的电阻为2.19×10-3Ω,铜内自由电子密度为8.5×1029m-3。求:
(1)通过铜棒的电流
(2)铜棒内的电场强度
(3)自由电子定向移动的速率。
(1)通过铜棒的电流
(2)铜棒内的电场强度
(3)自由电子定向移动的速率。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155558753677.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155558565501.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155558675733.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155558753677.png)
试题分析:(1)根据欧姆定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241555590491491.png)
(2)根据公式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155559221598.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241555593141432.png)
(3)单位长度内,含有电子个数为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241555594231486.png)
根据电流的微观表达式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155559657551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241555597671058.png)
点评:基础题,比较简单,关键是对公式的正确掌握
![](http://thumb2018.1010pic.com/images/loading.gif)
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